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The polynomial $f(x)$ has degree 3. If $f(-1) = 15$, $f(0)= 0$, $f(1) = -5$, and $f(2) = 12$, then what are the $x$-intercepts of the graph of $f$?

Guest Jul 31, 2018
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The polynomial $f(x)$ has degree 3. If
$f(-1) = 15$,
$f(0)= 0$,
$f(1) = -5$, and
$f(2) = 12$,
then what are the $x$-intercepts of the graph of $f$?

 

\(\begin{array}{|llcll|} \hline \boxed{f(x)=ax^3+bx^2+cx+d }\\\\ f(0) = 0: & 0 = a\cdot0^3+b\cdot 0^2+c\cdot 0 + d \\ & \mathbf{\boxed{d=0}} \\\\ \boxed{f(x)=ax^3+bx^2+cx+0 \\ f(x) = x(ax^2+bx+c) }\\\\ f(-1) = 15: & 15 = (-1)\left(a(-1)^2+b(-1)+c \right) \\ & 15 = (-1) (a-b+c ) \\ &\mathbf{ 15 = -a+b-c \qquad (1) } \\ f(1) = -5: & -5 = 1\cdot\left(a(1)^2+b(1)+c \right) \\ &\mathbf{ -5 = a+b+c \qquad (2) }\\\\ (1)+(2): & 15-5 =b+b \\ & 10 = 2b \\ & \mathbf{\boxed{b=5}} \\\\ \text{see }(1): & 15 = -a+b-c \quad b=5 \\ & 15 = -a+5-c \\ & 10 = -a-c \\ & \mathbf{c = -10-a \qquad (3)} \\\\ \boxed{f(x) = x(ax^2+bx+c) \\ f(x)=x(ax^2+5x-10-a)} \\\\ f(2) = 12: & 12 = 2\cdot(a\cdot2^2+5\cdot 2-10-a ) \\ & 6 = 4a-a \\ & 6 = 3a \\ & \mathbf{\boxed{a=2}} \\\\ \text{see }(3): & c = -10-a \quad a=2 \\ & c = -10-2 \\ & \mathbf{\boxed{c=-12}} \\ \hline \end{array}\)

 

\(\text{The polynomial $f(x)$ of degree $3$ is:} \\ \boxed{f(x) = 2x^3 + 5x^2-12x \\ f(x) = x(2x^2+5x-12) }\)

 

\(\text{x-intercepts $f(x) = 0\ ?$}\\ \begin{array}{|rcll|} \hline \boxed{ f(x) = x(2x^2+5x-12) }\\\\ 0 &=& x(2x^2+5x-12) \\ \mathbf{x_1} &\mathbf{=}&\mathbf{ 0} \\\\ 2x^2+5x-12 &=& 0 \\ x &=& \dfrac{-5\pm \sqrt{25-4\cdot 2\cdot (-12) } }{2\cdot 2} \\ x &=& \dfrac{-5\pm \sqrt{25+96 } }{4} \\ x &=& \dfrac{-5\pm \sqrt{121} }{4} \\ x &=& \dfrac{-5\pm 11 }{4} \\\\ x_2 &=& \dfrac{-5+ 11 }{4} \\ x_2 &=& \dfrac{6}{4} \\ \mathbf{x_2} & \mathbf{=}& \mathbf{ 1.5 } \\\\ x_3 &=& \dfrac{-5- 11 }{4} \\ x_3 &=& -\dfrac{16}{4} \\ \mathbf{x_3} & \mathbf{=}& \mathbf{ -4 } \\ \hline \end{array} \)

 

The x-intrcepts are: \( x=-4,\ x=0,\ x=1.5\)

 

The graph:

 

laugh

heureka  Jul 31, 2018

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